Class 11th

(5³)^740 * 5² = 5² (125)^740 = 5² (126 – 1)^740
So remainder = 4.

C.V. = (σ/x? ) * 100 ⇒ σ = (C.V. * x? ) / 100
∴ σ? = (50*30)/100 = 15 and σ? = (60*25)/100 = 15 ⇒ σ? - σ? = 0

Equation of tangent is (x/a) (1/2) + (y/b) (√3/2) = 1
Equation of auxiliary circle is x² + y² = a²
Homogenising (ii) with (i) and making coefficient of x² + coefficient of y² = 0
⇒ (3a²/4b²) - (7/4) = 0 ⇒ e = 2/√7

$N_1{V}_1=N_2{V}_2$

$0.01*V_1=0.01*20*3$

$V_1=60\mathrm{m}\mathrm{L}$

log? (abc) = 6 ⇒ abc = 6?
a = b/r & c = br
⇒ b = 36 and a = 36/r
⇒ r = 2,3,4,6,9,12,18
Also b - a = 36 (1 - 1/r) is a perfect cube.
∴ r = 4 ⇒ a + b + c = 9 + 36 + 144 = 189

β² – 6β + 12 = 0 β − 6 = -12/β
∴ Reqd. Expression is (α – 2)³)? + (-12)¹² / (αβ)¹²) - 1
= (α³ – 8 – 6α (α – 2)? + (12)¹² / (12)¹²) - 1 = (α (α² – 6α + 12) – 8)? = 8? = 2¹²

Solving, x² – 9 = kx² ⇒ x² (k − 1) + 9 = 0 ⇒ x? + x? = 0 and x? = 9 / (k-1)
|x? - x? | = 10 = √ (x? + x? )² - 4x? x? ) ⇒ k = 16/25

A: 3 numbers are in A.P.
B: 3 numbers are even
P (B/A) = n (A ∩ B) / n (A) = (²? C? + ²? C? ) / (? C? +? C? ) = (25 * 24) / (50 * 49) = 12/49

Kindly consider the following figure

As the difference in $3^{\text{rd}}$ ${}^{}$ and $4^{\text{th}}$ ${}^{}$ ionisation energy is high so atom contains 3 valence electrons. Hence, valency of the atom is 3 and the metal is trivalent. 4 mole of trivalent metal hydroxide will react with 6 mole of ${\mathrm{H}}_2{\mathrm{S}\mathrm{O}}_4$ ${}_{}{}_{}$, $\left(2\mathrm{M}(\mathrm{O}\mathrm{H}{)}_3+3{\mathrm{H}}_2{\mathrm{S}\mathrm{O}}_4?{\mathrm{M}}_2{\left({\mathrm{S}\mathrm{O}}_4\right)}_3+6{\mathrm{H}}_2\mathrm{O}\right)$